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Let's define the following function:

f[t_, x_] = t^2 + t*x*(1 - x)

such that $x \in (0,1)$ and `$t \in (0,0.02)$ The problem is when I want to define the following function:

z[t_,x_] = Interpolation[Table[{x, f[t, x] Piecewise[{{1, 0.2 < x < 0.8}}]}, {t, 0,0.02},{x, 0,1, 1/24}]];

This is not working!! (I think the problem is because I discretize in space and not in time!!)

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  • $\begingroup$ I'm not familiar with your goal but here are a couple of notes you probably missed: _ for arguments in defining f, using = instead of := in defining z, and also when you don't specify step in Table it's 1 which in your case t has almost no effect. Running Table alone will show you it will return a nested list not the kind of input Interpolation wants. Another thing is using similar names (z and Table). I think I should also remind using ClearAll before correcting the notes. $\endgroup$
    – Ben Izd
    Apr 1 at 11:44
  • $\begingroup$ You have right, the problem is that I get a list not the kind of input Interpolation wants. and I don't know how to solve this problem, I need a way to define a function of two parameters, t and x such that ${t, 0, 0.02}$ and ${x, 0,1, 1/24}$!! I don't know how!! $\endgroup$
    – walid fssm
    Apr 1 at 12:09

2 Answers 2

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I'm not 100% clear what you are trying to achieve. If you want to sample your continuous function $f(t,x)$ at only discrete values of $x$, but be able to call it later for any value of $t$ then:

f[t_,x_]:= (t^2 + t x (1-x)) Piecewise[{{1,0.2<x<0.8}}];

which we can sample then using:

zdis[t_]:= Table[f[t,x],{x,0,1,1/24}];

You can plot all values then using:

Plot[Evaluate[zdis[t]],{t,0,0.04},PlotLabels-> Table[InputForm[x], {x, 0, 1, 1/24}]]

giving:

Plots of zdis[t]

Addendum:

To get the $i$th value where $x = i/24$ for $i\in [0,24]$ you can either just call

zdis[t][[i+1]]

or build this into the original definition as per:

zdis[t_,i_?IntegerQ]:= f[t,i/24];

where the ?IntegerQ requires the input to be an integer.

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  • $\begingroup$ thank you, this is exactly what I'm looking for, I just need a little more detail please. after you define the table $zdis[t_]$, can you please show me how I can define a new function $g(t , i)$ such that for each value of $ t \in (0,0.02)$ and $i \in (0,1/24,1/12,....,1)$ I get the value !! $\endgroup$
    – walid fssm
    Apr 1 at 17:58
  • $\begingroup$ See the addendum to the answer. Note that the $i$ input in my answer is an integer, not a multiple of 1/24, which is harder to do (and unnecessay). Defining this secondary function zdis is not really necessary in the first place if you only ever call f[t,x] with values of $x \in [0,1/24,2/24,....]$. $\endgroup$
    – degenerate_matter
    Apr 1 at 18:38
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Perhaps FunctionInterpolationis what you're looking for.

Try

f[t_, x_] = t^2 + t*x*(1 - x)
fip = FunctionInterpolation[f[t, x] Piecewise[{{1, 0.2 < x < 0.8}}], {t, 0, 0.02}, {x, 0, 1 }] 
Plot3D[fip[t, x], {t, 0, 0.02}, {x, 0, 1}]

enter image description here

Addendum

For discrete interpolation in x we transform f[t,x] in a product {t^2,t}. {1,x(1-x).

Now it's possible to interpolate x:

ip = Interpolation[Table[{x, Piecewise[{{1, 0.2 < x < 0.8}}] {1, x (1 - x)}}, {x, 0, 1,1/24}],InterpolationOrder->1];

Plot[ip[x],{x,0,1}]

enter image description here

Final function follows to {t^2,t}.ip[x]!

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  • $\begingroup$ Thank you for your answer, But I see that you consider x in the whole interval (0,1). Me I need $(x, 0,1, 1/24)$. $\endgroup$
    – walid fssm
    Apr 1 at 12:13
  • $\begingroup$ That means you're looking for a discrete interpolation ,{x, 0,1, 1/24}? $\endgroup$
    – Ulrich Neumann
    Apr 1 at 12:21
  • $\begingroup$ that's right, I'm looking to define a discrete function in space but not in time.!! $\endgroup$
    – walid fssm
    Apr 1 at 12:24
  • $\begingroup$ @walidfssm See my modified answer, now with discrete interpolation x. $\endgroup$
    – Ulrich Neumann
    Apr 1 at 13:20
  • $\begingroup$ Thank you for your answer, But, this is not what I need. is it not possible to define it without separating it into a product!! $\endgroup$
    – walid fssm
    Apr 1 at 13:49

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